A359451 Decimal expansion of Sum_{k>=1} 1/A359450(k).

2, 0, 8, 6, 3, 7, 7, 6, 6, 5, 0, 0, 5, 9, 8, 8, 7, 1, 6, 0, 8, 9, 7, 5, 5, 8, 5, 6, 7, 3, 4, 1, 3, 2, 7, 7, 2, 6, 9, 2, 0, 2, 2, 0, 9, 6, 9, 2, 2, 3, 9, 5, 1, 6, 9, 5, 1, 2, 3, 8, 3, 8, 5, 7, 9, 2, 1, 5, 3, 2, 0, 0, 0, 2, 8, 2, 1, 0, 0, 2, 6, 1, 4, 7, 1, 6, 0, 5, 8, 4, 8, 5, 2, 6, 7, 0, 9, 4, 9, 0, 7

Offset: 1

Author

Amiram Eldar, Jan 02 2023

Keywords

• nonn
• cons
• base

Comments

The problem of calculating the sum of this series was proposed by David Smith in Bornemann et al. (2004). The value that is given here is from his solution on the web page of this book. He shows that the series is slowly converging. E.g., the sum of the first 2^2000 - 1 terms is 1.95403... .

Examples

			2.08637766500598871608975585673413277269202209692239...
		

References

• Daniel D. Bonar and Michael J. Khoury, Jr., Real infinite Series, The Mathematical Association of America, 2006, pp. 159, 190-191.
• Kiran S. Kedlaya, Daniel M. Kane, Jonathan M. Kane, and Evan M. O'Dorney, The William Lowell Putnam Mathematical Competition 2001-2016: Problems, Solutions, and Commentary, American Mathematical Society, 2020, pp. 86-87.

Links

• Folkmar Bornemann, Dirk Laurie, Stan Wagon, and Jörg Waldvogel, The SIAM 100-Digit Challenge, A Study in High-Accuracy Numerical Computing, SIAM, Philadelphia, 2004. See Appendix D, Problem 2, p. 281.
• David Smith, On a slowly converging sum, Solution to Problem 2, Loyola Marymount University, Note of November 2003.

Crossrefs

Cf. A001008, A002805, A359450.

Formula

Equals 5/3 + Sum_{k>=3} (H(2^k-1)-H(2^(k-1)-1))/A359450(k), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number.

Equals (1/(1-log(2))) * (5/3 - 3*log(2)/2 + Sum_{k>=3} (H(2^k-1)-H(2^(k-1)-1)-log(2))/A359450(k)).

Both formulas are from Smith (2003).